GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 24 Sep 2018, 22:59

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M12-13

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

16 Sep 2014, 00:46
00:00

Difficulty:

85% (hard)

Question Stats:

56% (01:20) correct 44% (01:50) wrong based on 126 sessions

### HideShow timer Statistics

What is the equation of the line $$k$$ that is perpendicular to line $$y = 2x$$ and passes through point $$(a, b)$$?

(1) $$a = -b$$

(2) $$a - b = 1$$

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

16 Sep 2014, 00:46
Official Solution:

Notice that we can get the equation of line $$k$$ which is perpendicular to line $$y=2x$$ if we know ANY point that line $$k$$ passes through. So, to get the equation of line $$k$$ we need the values of $$a$$ and $$b$$.

(1) $$a = -b$$. Not sufficient.

(2) $$a - b = 1$$. Not sufficient.

(1)+(2) $$a = -b$$ and $$a - b = 1$$, we have two distinct linear equations with two unknowns so we can solve for $$a$$ and $$b$$. Sufficient.

_________________
Current Student
Joined: 12 Aug 2015
Posts: 287
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

24 Nov 2015, 01:53
hi Bunuel

can't A suffice to find solution? y=2x is given as going from quad 3 through origin to quad 1 with slope 2.

slope of the perped line is -1/2, relatioship between points is given as whenever a is positive the b bears the same value but with opposite sign (10,-10 or -10;10) - hence the second line has to go through the origin from quad 2 to quad 4?

thanks
_________________

KUDO me plenty

Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

24 Nov 2015, 10:24
hi Bunuel

can't A suffice to find solution? y=2x is given as going from quad 3 through origin to quad 1 with slope 2.

slope of the perped line is -1/2, relatioship between points is given as whenever a is positive the b bears the same value but with opposite sign (10,-10 or -10;10) - hence the second line has to go through the origin from quad 2 to quad 4?

thanks

Why should line k go through the origin?
_________________
Intern
Joined: 29 Jun 2016
Posts: 11

### Show Tags

02 Jul 2016, 03:38
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

02 Jul 2016, 05:17
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

Check the diagram below:

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.

_________________
Senior Manager
Joined: 31 Mar 2016
Posts: 395
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

### Show Tags

24 Jul 2016, 00:34
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Bunuel,

If the intention is merely to find values of a and b then what is the significance of "Perpendicularity" here? Please explain
Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

24 Jul 2016, 02:42
Senthil7 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Bunuel,

If the intention is merely to find values of a and b then what is the significance of "Perpendicularity" here? Please explain

The question asks to find the equation of k that is perpendicular to line y=2x and passes through point (a,b). We need to find the values of a and b to answer the question.
_________________
Intern
Joined: 31 May 2016
Posts: 4

### Show Tags

16 Aug 2016, 13:36
I think that this is a high-quality question and the explanation is very clear. Thanks for posting.
Intern
Joined: 11 Aug 2016
Posts: 1

### Show Tags

17 Aug 2016, 06:30
S1) take different values for a,b that meet the given statement (a=-b) and form lines equations of lines with slope -1/2. You can see that for different values of (a,b), you get different equations.

S2) same as statement 1

Both) Both equations give a point a=1/2 and b=-1/2. Which hivea only one equation 2x+4y=-1

So the ans is C. Hope fully my logic ia correct.

Posted from my mobile device
Senior Manager
Joined: 12 Mar 2013
Posts: 283

### Show Tags

07 Oct 2016, 13:30
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

Check the diagram below:

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png

i want to apologize, as i am very very weak at coordinate geometry.

First, please give a suggestion how can i learn coordinate geometry.

2nd, isn't it obvious in option A that b=5, a= -5; b= 1 a= -1; b=0, a=0; b=1, a= -1?

sorry again if i ask something foolish
_________________

We Shall Overcome... One day...

Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

08 Oct 2016, 03:16
nahid78 wrote:
Bunuel wrote:
subhajit1 wrote:
Hi friends,

the line y=2x surely passes through the origin as the coordinates (0,0) satisfies the equation.

Now coming to the point, the solution says both the statements are required, but if you take the slope of second line as -1/2 then the equation for is

-1/2=(y-b)/(x-a)
or, 2y= -x +(a+b)
or, y=(-1/2)x+(a+b)/2

Now, applying statement 1 (a=-b)to this equation;

y=(-1/2)x+{a+(-a)}/2
or, y=(-1/2)x+ 0/2
or, y=(-1/2)x

Check the diagram below:

As you can see if a = 1 and b = -1 we have different line perpendicular to y = 2x than if a = 2 and b = -2.

Attachment:
Untitled.png

i want to apologize, as i am very very weak at coordinate geometry.

First, please give a suggestion how can i learn coordinate geometry.

2nd, isn't it obvious in option A that b=5, a= -5; b= 1 a= -1; b=0, a=0; b=1, a= -1?

sorry again if i ask something foolish

For coordinate geometry check this: math-coordinate-geometry-87652.html

For more on this question check here: what-is-the-equation-of-the-line-k-that-is-perpendicular-to-line-y-69481.html
_________________
Intern
Joined: 26 Jul 2016
Posts: 3

### Show Tags

20 Dec 2016, 06:23
Hi Bunuel,

A little clarification
equation of line perpendicular to y-2x=0 will be -2y-x+c=0 ??
In order to find equation of a line perpendicular to ax+by+c=0.....interchange coefficients of x and y and change the sign in between....so equation of line perpendicular to ax+by+c=0 will be bx-ay+k=0.
Is this approach correct?
Intern
Joined: 27 Oct 2015
Posts: 19

### Show Tags

27 Dec 2016, 11:18
I think this is a high-quality question and I agree with explanation.
Current Student
Joined: 12 Oct 2015
Posts: 3

### Show Tags

14 Jan 2017, 10:26
Hi Bunuel,

If the question were to be a slightly different one, would the below Statement be sufficient?

Question stem: Find equation of the line K, perpendicular to a line y= -1/2x and passes through point (a,b).

statement 1) 2a=b

I understood the point that given a slope + either the y-intercept or a specific point, we can plot the specific line. Where I am getting confused is whether we can still identify the line by having the slope and a generic equation such as 2a=b.

When I tried plotting several lines with the slope = 2 and with different y-intercepts, I could not come up with any point that satisfy 2a=b and any other lines except y=2x+0.

Intern
Joined: 16 Jan 2017
Posts: 6

### Show Tags

17 Apr 2017, 02:15
I do not understand why we cannot say that stat I implies a=b=0, hence being sufficient while statement II implies basically all consecutive integers, not being sufficient. Can someone please explain?
Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

17 Apr 2017, 02:20
Kontaxis wrote:
I do not understand why we cannot say that stat I implies a=b=0, hence being sufficient while statement II implies basically all consecutive integers, not being sufficient. Can someone please explain?

a = -b does not necessarily means that a = b= 0. Why not a= 1 and b = -1?
_________________
Intern
Joined: 16 Jul 2017
Posts: 1

### Show Tags

10 Sep 2017, 22:59
Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?
Math Expert
Joined: 02 Sep 2009
Posts: 49438

### Show Tags

10 Sep 2017, 23:36
RekhaKulkarni wrote:
Hi, Can anyone tell me what will be the final equation of the line once we know what a and b are?

From a=-b and a-b=1, we can get that a = 1/2 and b = -1/2. So, we know that k passes through the point (1/2, -1/2) and is perpendicular to line y = 2x (slope = 2)

The two lines are perpendicular if and only if the product of their slopes is -1, so m*2 = -1 and m = -1/2 (the slope of line k).

Finally, the equation of a straight line that passes through a point $$P_1(x_1, y_1)$$ with a slope m is: $$y-y_1=m(x-x_1)$$. Substitute: $$y-(-\frac{1}{2})=-\frac{1}{2}(x-\frac{1}{2})$$ --> $$y = -\frac{x}{2} - \frac{1}{4}$$.

Check below:

For Coordinate Geometry check:

24. Coordinate Geometry

For more check:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.

Attachment:
Untitled.png

>> !!!

You do not have the required permissions to view the files attached to this post.

_________________
Intern
Joined: 30 Apr 2017
Posts: 4

### Show Tags

13 Sep 2017, 23:20
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i read the whole thread, and still didn't understand the explanation. I do understand that we don't need to find the value of a and b, question is why would we discard 1） and 2) as insufficient, what's the reasoning behind it.
Re M12-13 &nbs [#permalink] 13 Sep 2017, 23:20

Go to page    1   2    Next  [ 26 posts ]

Display posts from previous: Sort by

# M12-13

Moderators: chetan2u, Bunuel

## Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.